Calculating Force given velocity and mass of two objects

I am trying to calculate the force of a 36,000kg truck traveling with a constant velocity of 26.8224 m/s colliding with a 62kg person who is standing still (0 m/s velocity). I know that ∑F=0 ∴ Ftruck→person=-Fperson→truck, that F=ma, and a=v/t. But this would be instantaneous and as I recall that would make it important to calculate impulse and/or momentum. I know that momentum is p=mv and impulse is J=Δp. This should be easy as I have done plenty of these kinds of problems in school, but right now I seem to be brain farting all this. I can't seem to relate the given terms and the unknowns to solve for F. :s I'm trying to use it in an example for someone else so I can relate the amount of force exerted on an object in LEO to something like getting hit by a truck. Right now I have the gravitational force exerted by the Earth on the ISS being 7.94072 × 10^8 N. I want to tell them "that is like being hit by [x amount] of trucks."

Do you want to treat this as an elastic collision with Superman, or the slightly messier inelastic collision of the real world?

Preferably inelastic so I can provide a more accurate analogy.

It's kind of embarrassing to not remember this. I did it just a year ago in Physics I, but Physics II (which I just finished) wasn't about mechanics and the Engineering Mechanics class I just took was in Static Equilibrium this semester so my brain is trying to calculate moments and charges. -_______-;;

So, figure the least dimension of this individual is ~ 0.3 m. In the time it takes the truck to travel 0.3 m, the roadkill will be accelerated to the post-collision velocity of (36k x 26.8224 m/s)/(36,062 kg (let's not get too messy --- we'll conserve the mass) ). Got it from here?

So, figure the least dimension of this individual is ~ 0.3 m. In the time it takes the truck to travel 0.3 m, the roadkill will be accelerated to the post-collision velocity of (36k x 26.8224 m/s)/(36,062 kg (let's not get too messy --- we'll conserve the mass) ). Got it from here?

I'm sorry, did you mean (36,000 kg x 26.8224 m/s)/36,062 kg? Essentially I will be left with the velocity of the (now dead) person and I would use the acceleration as a function of time with a=v2/d with the distance being 0.3m right? Then I would plug that into the force equation, but would I use the mass of the truck, mass of the person, or net mass of the system?

Inelastic collision --- momentum is conserved. Yes, you have accelerated the corpse to the final velocity during the travel time for the 0.3 m, using just the mass of the late unfortunate. Or, you can look at the force decelerating the truck over that same distance from vinitial to vfinal, in which case you use the mass of the truck.

Inelastic collision --- momentum is conserved. Yes, you have accelerated the corpse to the final velocity during the travel time for the 0.3 m, using just the mass of the late unfortunate. Or, you can look at the force decelerating the truck over that same distance from vinitial to vfinal, in which case you use the mass of the truck.

Thank you so much. You helped me complete my explanation of orbital mechanics with respect to LEO with a lovely analogy to round it out. I explained that zero G does not mean 0 Fg in this sense. And you helped me calculate that the Earth pulls on the ISS with a gravitational force equal to that of 5,359 semi-trucks hitting the average human. It was so good some of my friends are tweeting NASA to hire me. Lol.